题目内容
【题目】化简:(1)–3x2y+2x2y+3xy2–xy2;
(2)4x2–(2x2+x–1)+(2–x2+3x).
【答案】(1) –x2y+2xy2;(2) x2+2x+3.
【解析】
(1)把同类项进行合并即可得;
(2)先去括号,然后再合并同类项即可得答案.
(1)–3x2y+2x2y+3xy2–xy2
=(-3+2)x2y+(3-1)xy2
=–x2y+2xy2;
(2)4x2–(2x2+x–1)+(2–x2+3x)
=4x2–2x2–x+1+2–x2+3x
=x2+2x+3.
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